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Suppose `A` and `B` are two non singular matrices such that `B != I, A^6 = I` and `AB^2 = BA`. Find the least value of `k` for `B^k = 1`A.B.C.D. |
Answer» Given, `AB = BA^(2) rArr B= A^(-1) BA^(2) rArr B^(3) = I` `rArr (A^(-1) BA A) ( A^(-1) BA A ) (A^(-1) BA A) = I` `rArr (A^(-1) BA A) (BA ) ( BA A) = I [because A^(-1) A=I]` `rArr A^(-1) B (BA^(2)) (BA^(2)) A A=I` `rArr A^(-1) B(BA^(2)) (BA^(2)) A A = I [ because AB =BA^(2) ] ` `rArr A^(-1) BBA (AB) A^(4) = I` `rArr A^(-1) BBA (BA^(2) )A^(4) = I [because AB = BA^(2)]` `rArr A^(-1) BB (AB) A^(6) = I` `rArr A^(-1) BB(BA^(2)) A^(6) = I [because AB = BA^(2)]` `rArr A^(-1) B^(3) A^(8) = I` ` rArr (A^(-1)I) A^(8) = I [ B^(3) = I]` `rArr A^(-1) A^(8) = I` `rArr A^(7) = I = A^(k) [ because A^(k) = I]` `rArr A^(k) = A^(7)` `therefore` Least value of `k` is 7. |
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